Answer:
r is a straight line
completing or enhancing something.
Step-by-step explanation:
Answer:
a) 
b) 
And replacing we got:
![P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-%20%5B0.2%2B0.3%2B0.1%5D%3D%200.4)
c) 
d) 
e) 
f) 
And replacing we got:

And the variance would be:
![Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4](https://tex.z-dn.net/?f=%20Var%28X0%20%3DE%28X%5E2%29-%20%5BE%28X%29%5D%5E2%20%3D%206.4%20-%282%5E2%29%3D%202.4)
And the deviation:

Step-by-step explanation:
We have the following distribution
x 0 1 2 3 4
P(x) 0.2 0.3 0.1 0.1 0.3
Part a
For this case:

Part b
We want this probability:

And replacing we got:
![P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%203%29%20%3D%201-%20%5B0.2%2B0.3%2B0.1%5D%3D%200.4)
Part c
For this case we want this probability:

Part d

Part e
We can find the mean with this formula:

And replacing we got:

Part f
We can find the second moment with this formula

And replacing we got:

And the variance would be:
![Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4](https://tex.z-dn.net/?f=%20Var%28X0%20%3DE%28X%5E2%29-%20%5BE%28X%29%5D%5E2%20%3D%206.4%20-%282%5E2%29%3D%202.4)
And the deviation:

D 24 do order of operations perenthesis, exponents, multiplication or division, addition or subtraction
Y intercept is (0,-12), the x-intercept is (4,0)
21 friends.
44 - 2x = 2, where x is the number of friends Kristin has.
Subtract 2 on each side.
42 - 2x = 0
Add 2x on each side.
42 = 2x
Divide by 2 on each side.
42 / 2 = x
21 = x
Kristin has 21 friends.
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